Integers are positive or negative whole numbers without a decimal or fractional component. Multiplying and dividing two or more integers is not very different from multiplying and dividing basic whole numbers. The key difference is that, because some integers are negative, you must keep track of their signs. Taking your integers' signs into account, you may proceed by multiplying as normal.

Steps

General Information

1

Know your integers. An integer is any whole number that can be represented without using a fraction or decimal. Integers can be positive, negative, or zero. For instance, the following numbers are integers: 1, 99, -217, and 0. However, these numbers are not: -10.4, 6 ¾, 2.12.

Absolute values can be integers, but they are not necessarily. An absolute value of any number is the "size" or "amount" of the number, regardless of its sign. Another way to put this is that a given number's absolute value is that number's distance from zero. So, the absolute value of an integer is always an integer. For example, the absolute value of -12 is 12. The absolute value of 3 is 3. The absolute value of 0 is 0.

The absolute values of numbers that are not integers, however, will never be integers. For example, the absolute value of 1/11 is 1/11 - a fraction, and therefore not an integer.

2

Know your basic times tables. The process of multiplying or dividing integers, whether they're large or small, is much, much quicker and easier if you have memorized the products of every pair of numbers from 1 to 10. This information is usually referred to in school as "times tables". As a refresher, below is a basic 10X10 times table. The numbers across the top and left side of the table list the numbers from 1 to 10. To find the product of two of these numbers, find the cell where the row and column of your two desired numbers intersect:

Times table from 1 to 10.

1

2

3

4

5

6

7

8

9

10

1

1

2

3

4

5

6

7

8

9

10

2

2

4

6

8

10

12

14

16

18

20

3

3

6

9

12

15

18

21

24

27

30

4

4

8

12

16

20

24

28

32

36

40

5

5

10

15

20

25

30

35

40

45

50

6

6

12

18

24

30

36

42

48

54

60

7

7

14

21

28

35

42

49

56

63

70

8

8

16

24

32

40

48

56

64

72

80

9

9

18

27

36

45

54

63

72

81

90

10

10

20

30

40

50

60

70

80

90

100

Method1

Multiplying Integers

1

Count the number of negative signs in your multiplication problem. A basic multiplication problem between two or more positive numbers will always result in a positive answer. However, each negative sign added to a multiplication problem flips the sign from positive to negative or vice versa. To begin an integer multiplication problem, count the number of negative signs in the problem.

Let's use the example problem -10 × 5 × -11 × -20. In this problem, we can clearly see three negative signs. We'll use this information in the next step.

2

Decide the sign of your answer based on the number of negative signs in the problem. As noted above, the answer to a multiplication problem involving only positive integers will be positive. For each negative negative sign in your problem, flip the sign of your answer. In other words, if your problem has one negative sign, your answer will be negative; if it has two, your answer will be positive, and so on. A good rule of thumb is that odd numbers of negative signs give negative answers and even numbers of negative signs give positive answers.

In our example, we have three negative signs. Three is an odd number, so we know our answer is negative. We can put a negative sign in the space for our answer, like this: -10 × 5 × -11 × -20 = -__

3

Multiply numbers from 1 - 10 using basic times table knowledge. The product of any two numbers lesser than or equal to 10 is covered in basic times tables (see above). For these simple cases, just write the answer. Remember that, in problems that only use multiplication signs, you can move the integers around so that you're able to multiply simple numbers with each other.

In our example, 10 × 5 is covered in the basic times table. We don't have to account for the negative sign on the ten because we've already found the sign of our answer. 10 × 5 = 50. We can insert this into our problem like this: (50) × -11 × -20 = -__

If necessary, break larger numbers into manageable chunks. If your multiplication problem involves numbers greater than ten, you don't necessarily have to use long multiplication. First, see if you can break one or more of your numbers down into smaller, more workable pieces. Since, with basic times table knowledge, you can solve simple multiplication problems almost instantly, breaking a difficult problem into several of these easy problems is usually simpler than solving the single difficult problem.

Let's look at the second half of our example problem, -11 × -20. We can omit the signs because we've already figured out the sign of our answer. 11 × 20 looks intimidating, but if we rewrite the problem as 10 × 20 + 1 × 20, suddenly, it's much more manageable. 10 × 20 is just 2 times 10 × 10, or 200. 1 × 20 is just 20. Adding up our answers, we get 200 + 20 = 220. We can re-insert this into our problem as follows: (50) × (220) = -__

5

For more difficult numbers, use long multiplication. If your multiplication problem involves two or more numbers greater than 10 and you're not able to find the answer by dividing your problem into workable chunks, you can still solve via long multiplication. In long multiplication, you line your answers up as you would in an addition problem and multiply each digit in the bottom number by each digit in the top number. If the bottom number has more than one digit, you'll need to account for digits in the tens, hundreds, and so on place by adding zeroes to the right side of your partial answer. Finally, to get your final answer, add up all the partial answers.

Let's return to our example problem. Now, we must multiply 50 by 220. This will be difficult to break into easier chunks, so let's use long multiplication. Long multiplication problems are easier to keep track of if the smaller number is on the bottom, so let's write our problem with 220 on top and 50 on the bottom.

First multiply the digit in the ones place of the bottom number by each digit of the top number. Since 50 is on the bottom, 0 is the digit in the ones place. 0 × 0 is 0, 0 × 2 is 0, and 0 × 2 is zero. In other words, 0 × 220 is zero. Write this below your long multiplication problem in the ones place. This is our first partial answer.

Next, we'll multiply the digit in the tens place of our bottom number by each digit of the top number. 5 is the digit in the tens place of 50. Since this 5 is in the tens place, rather than the ones place, we write a zero below our first partial answer in the ones place before proceeding. Next, we multiply. 5 × 0 is 0. 5 × 2 is 10, so write 0 and add one to the product of 5 and the next digit. 5 × 2 is 10. Normally, we would write 0 and carry the 1, but in this case we also add the 1 from the previous problem, giving us 11. Write down "1". Carrying the 1 from the tens place of 11, we see that we're out of digits, so we just write it to the left of our partial answer so far. Recording all this, we're left with 11,000.

Next, we just add. 0 + 11,000 is 11,000. Since we know the answer to our original problem is negative, we can safely say that -10 × 5 × -11 × -20 = -11,000.

Method2

Dividing Integers

1

As before, decide the sign of your answer based on the number of negative signs in the problem. Introducing division to a math problem doesn't change the rules regarding negative signs. If there are an odd number of negative signs, the answer is negative, while if there are an even number of negative signs (or none at all) the answer will be positive.

Let's use an example problem with both multiplication and division. In the problem -15 × 4 ÷ 2 × -9 ÷ -10, there are three negative signs, so the answer will be negative. As before, we can put a negative sign in the space for our answer, like this: -15 × 4 ÷ 2 × -9 ÷ -10 = -__

2

Make simple divisions using your multiplication knowledge. Division can be thought of as multiplication done backwards. When you divide one number by another, you're asking in a roundabout way, "how many times does the second number fit into the first?" or, in other words, "what do I need to multiply the second number by to get the first?" See the basic 10 x 10 times table for reference - if you're asked to divide one of the answers in the times table by any number n from 1 - 10, you'll know that the answer is just the other number from 1 - 10 needed to multiply n to get it.

Use long division when necessary. As with multiplication, when you come across a division problem that's too difficult to work out mentally or with a times table, you have the option of solving with a long-form approach. In a long division problem, you write your two numbers in a special sideways L-shaped bracket, then divide digit-by-digit, shifting your partial answers to the right as you go to account for the decreasing value of the digits you're dividing - hundreds, then tens, then ones, and so on.

Let's use long division in our example problem. We can simplify -15 × (2) × -9 ÷ -10 to 270 ÷ -10. We'll ignore the signs as usual because we know the sign of our final answer. Write 10 to the left of the L-shaped bracket and write 270 underneath it.

We begin by dividing the first digit of the number under the bracket by the number to the side. The first digit is 2 and our number to the side is 10. Since 10 doesn't fit in two, we'll instead use the first two digits. 10 does fit in 27 - it fits in twice. Write "2" above the 7 under the bracket. 2 is the first digit in your answer.

Next, multiply the number to the left of the bracket by the digit you just discovered. 2 × 10 is 20. Write this under the first two digits of the number under the bracket - in this case, 2 and 7.

Subtract the numbers you just wrote. 27 minus 20 is 7. Write this at the bottom of your growing problem.

Drop down the next digit of the number under the bracket. This next digit of 270 is 0. Drop this down next to the 7 to make 70.

Divide your new number. Next, divide 10 into 70. 10 fits exactly 7 times into 70, so write at the top next to the 2. This is the second digit of your answer. Your final answer is 27.

Note that, in the event that 10 didn't divide evenly into our final number, we'd need to account for the amount of 10 that is left over - the remainder. For instance, if our final act was to divide 71, rather than 70, by 10, we would notice that 10 doesn't fit exactly in 71. It fits in 7 times, but there is 1 left over. In other words, we can fit seven 10s and an extra 1 in 71. We would write our answer, then, as "27 remainder 1" or "27 r1".